Toán $(x+1)\sqrt[]{x^{2}-2x+3}=x^{2}+1$ Tìm x 17/07/2021 By Lydia $(x+1)\sqrt[]{x^{2}-2x+3}=x^{2}+1$ Tìm x
Giải thích các bước giải: $(x+1)\sqrt{x^{2}-2x+3}=x^{2}+1$ $\Rightarrow(x+1)^2(x^{2}-2x+3)=(x^{2}+1)^2$ `=>(x^2+2x+1)(x^{2}-2x+3)=(x^{2}+1)^2` `=>x^4-2x^3+3x^2+2x^3-4x^2+6x+x^2-2x+3=x^4+2x^2+1` `=>x^4+4x+3=x^4+2x^2+1` `=>4x+3=2x^2+1` `=>2x^2+1-4x-3=0` `=>2x^2-4x-2=0` `=>(2x^2-4x+2)-4=0` `=>2(x^2-2x+1)-4=0` `=>2(x-1)^2=4` `=>(x-1)^2=2` `=>`\(\left[ \begin{array}{l}x-1=\sqrt2\\x-1=-\sqrt2\end{array} \right.\) `=>`\(\left[ \begin{array}{l}x=\sqrt2+1\\x=1-\sqrt2\end{array} \right.\) Trả lời
` (x + 1)\sqrt{x^2 – 2x + 3} = x^2 + 1 ` ` <=> (x + 1)^{2}(x^2 – 2x + 3) = x^4 + 2x^2 + 1 ` ` <=> (x^2 + 2x + 1)(x^2 – 2x + 3) = x^4 + 2x^2 + 1 ` ` <=> x^4 – 2x^3 + 3x^2 + 2x^3 – 4x^2 + 6x + x^2 – 2x + 3 = x^4 + 2x^2 + 1 ` ` <=> 3x^2 – 4x^2 + 6x + x^2 – 2x + 3 = 2x^2 + 1 ` ` <=> 4x + 3 = 2x^2 + 1 ` ` <=> 4x + 3 – 2x^2 – 1 = 0 ` ` <=> -2x^2 + 4x + 2 = 0 ` ` <=> -2(x^2 – 2x – 1) = 0 ` ` <=> x^2 – 2x – 1 = 0 ` ` <=> x = \frac{-(-2) ± \sqrt{(-2)^2 – 4.1.(-1)}}{2.1} ` ` <=> x = \frac{2 ± \sqrt{8}}{2} ` ` <=> x = \frac{2 ± 2\sqrt{2}}{2} ` ` <=> x = 1 ± \sqrt{2} ` ` <=> ` \(\left[ \begin{array}{l}x_1=1+\sqrt{2}\\x_2=1-\sqrt{2}\end{array} \right.\) Vậy ` S = {1 + \sqrt{2} ; 1 – \sqrt{2}} ` Trả lời
Giải thích các bước giải:
$(x+1)\sqrt{x^{2}-2x+3}=x^{2}+1$
$\Rightarrow(x+1)^2(x^{2}-2x+3)=(x^{2}+1)^2$
`=>(x^2+2x+1)(x^{2}-2x+3)=(x^{2}+1)^2`
`=>x^4-2x^3+3x^2+2x^3-4x^2+6x+x^2-2x+3=x^4+2x^2+1`
`=>x^4+4x+3=x^4+2x^2+1`
`=>4x+3=2x^2+1`
`=>2x^2+1-4x-3=0`
`=>2x^2-4x-2=0`
`=>(2x^2-4x+2)-4=0`
`=>2(x^2-2x+1)-4=0`
`=>2(x-1)^2=4`
`=>(x-1)^2=2`
`=>`\(\left[ \begin{array}{l}x-1=\sqrt2\\x-1=-\sqrt2\end{array} \right.\) `=>`\(\left[ \begin{array}{l}x=\sqrt2+1\\x=1-\sqrt2\end{array} \right.\)
` (x + 1)\sqrt{x^2 – 2x + 3} = x^2 + 1 `
` <=> (x + 1)^{2}(x^2 – 2x + 3) = x^4 + 2x^2 + 1 `
` <=> (x^2 + 2x + 1)(x^2 – 2x + 3) = x^4 + 2x^2 + 1 `
` <=> x^4 – 2x^3 + 3x^2 + 2x^3 – 4x^2 + 6x + x^2 – 2x + 3 = x^4 + 2x^2 + 1 `
` <=> 3x^2 – 4x^2 + 6x + x^2 – 2x + 3 = 2x^2 + 1 `
` <=> 4x + 3 = 2x^2 + 1 `
` <=> 4x + 3 – 2x^2 – 1 = 0 `
` <=> -2x^2 + 4x + 2 = 0 `
` <=> -2(x^2 – 2x – 1) = 0 `
` <=> x^2 – 2x – 1 = 0 `
` <=> x = \frac{-(-2) ± \sqrt{(-2)^2 – 4.1.(-1)}}{2.1} `
` <=> x = \frac{2 ± \sqrt{8}}{2} `
` <=> x = \frac{2 ± 2\sqrt{2}}{2} `
` <=> x = 1 ± \sqrt{2} `
` <=> ` \(\left[ \begin{array}{l}x_1=1+\sqrt{2}\\x_2=1-\sqrt{2}\end{array} \right.\)
Vậy ` S = {1 + \sqrt{2} ; 1 – \sqrt{2}} `